6.01SC Introduction to EECS I, Midterm 2 Solutions

6.01 Midterm 2 Name: Solutions Section: These solutions do not apply for the conflict exam. Enter all answers in the boxes provided. Clearly written work will be graded for partial credit. During the exam you may: • read any paper that you want to • use a calculator You may not • use a computer, phone or music player For staff use: 1. /12 2. /12 3. /12 4. /12 5. /18 6. /12 7. /12 8. /10 total: /100 1 1 Find the Voltage and Current (12 points). Determine V and I in the following circuit. I 15V V= I= 3Ω 10A 2Ω V 18V −1A Use superposition. If the current source is set to zero (replaced with an open circuit), then V would be 2Ω/(2Ω + 3Ω) × 10V= 6V and I would be 15V/(3Ω + 2Ω) = 3A. If the voltage source is set to zero (replaced by a short circuit), then V would be 10A×(2Ω||3Ω) = 10 × (2 × 3)/(2 + 3) = 12V and I would be −2Ω/(2Ω + 3Ω) × 10A= −4A. The sums are V = 6 + 12 = 18V and I = 3 − 4 = −1A. 2 2 Find the Resistance (12 points). Find the value of R so that Vo = 30V. 3Ω 15V R 10A Vo Enter your answer below, or enter none if no such value of R can be found. R= 6Ω By Ohm’s law, the current through R is IR = Vo . R Applying KCL to the node where R joins the current source and the 3Ω resistor, IR = 10A − Vo − 15V . 3Ω Equating these, and setting Vo = 30V yields R = 6Ω. 3 3 LTI SM (12 points). Write a difference equation for each of these machines if it describes an LTI system or give a very brief reason why it does not. The input to the machine at step n is x[n] and the output of the machine at step n is y[n]. class MM1(sm.SM): startState = [0, 0] def getNextValues(self, state, inp): return ([state[1], inp], 2*state[0]) y[n] = 2x[n − 2] class MM2(sm.SM): startState = [0] def getNextValues(self, state, inp): return (state + [inp], sum(state)) y[n] = y[n − 1] + x[n − 1] class MM3(sm.SM): startState = 0 def getNextValues(self, state, inp): return (max(state, inp), max(state, inp)) Max is not a linear operator. class MM4(sm.SM): startState = 0 def getNextValues(self, state, inp): return (state + 1, state) It is tempting to write: y[n] = y[n − 1] + 1 but the constant term is not legal in a difference equation. 4 4 Op-Amp Circuit (12 points). Determine Vo in the following circuit. Assume that the op-amp is ideal. 12Ω 15Ω 10V 6Ω Vo 5V 10Ω 6V Vo = Express the left voltage source and two left-most resistors as a Thevenin equivalent, with 10 Thevenin voltage 10+15 10V= 4V and Thevenin resistance 10Ω||15Ω = 10 × 15/(10 + 15) = 6Ω. 12Ω 6Ω 4V 6Ω 5V Vo Since V− = V+ , V− = 5V. So there must be 1/12A flowing left through the two 6Ω resistors. There must be a corresponding 1/12A flowing to the left through the 12Ω resistor, since no current enters the V− input of the op-amp. Vo is then the sum of V− = 5V and the 1V across the 12Ω resistor. 5 5 Run Length (18 points). One simple approach to sequence compression is called run-length encoding (RLE). A run is a subsequence of repeated entries. The idea is to represent the original sequence by a list of pairs of the form: (runLength, entry) For example, we could represent this list of digits: [3, 3, 3, 3, 5, 5, 9, 9, 9, 3, 3] by this: [(4, 3), (2, 5), (3, 9), (2, 3)] This representation is useful when there are likely to be long subsequences of repeated entries in the sequence. In this problem, you will define a class to represent and manipulate RLE sequences. class RLE: def __init__(self, seq): self.rleSeq = self.encode(seq) def encode(self, seq): # code 1 def decode(self): # code 2 def add(self, other): # code 3 6 5.1 Encoding Write the definition of the encode method, which takes a list of digits and returns an RLE-encoded list. def encode(self, seq): rle = [] prev = None count = 0 for x in seq: if x == prev: count = count+1 else: if prev: rle.append((count, prev)) prev = x count = 1 if prev: rle.append((count, prev)) return rle 7 5.2 Decoding Write the definition of the decode method, which returns a list of digits corresponding to the RLE-encoded list for the class instance. def decode(self): seq = [] for (count, entry) in self.rleSeq: for i in xrange(count): seq.append(entry) return seq 8 5.3 Addition Let’s define addition on our sequences as component-wise addition. Assume that both sequences are the same number of characters when decoded. >>> RLE([2,3,4,4,4]).add(RLE([2,3,3,3,4])) should produce a new instance of the RLE class whose content is: [(1, 4), (1, 6), (2, 7), (1, 8)] Don’t try to be efficient in your solution. It’s fine to decode the sequences to add them. def add(self, other): seq1 = self.decode() seq2 = other.decode() return RLE([x + y for (x, y) in zip(seq1, seq2)]) 9 6 Make it Equivalent (12 points). Determine values of R1 and R2 in the following circuit R2 10A R1 so that • the Thevenin equivalent voltage VT = 1V, and • the Thevenin equivalent resistance RT = 1Ω. R1 = 1 Ω 10 R2 = 9 Ω 10 The Thevenin voltage is the open-circuit voltage: VT = R1 × 10A = 1V. Thus R1 = 1 10 Ω. The Thevenin resistance is VT over the short-circuit current: RT = 1V R1 R1 +R2 10A Solving, we get R2 = = 1Ω. 9 10 Ω. 10 7 Current from Current Sources (12 points) Determine an expression for Io in the following circuit. Io I1 R3 R1 R2 I2 R1 I1 − R2 I2 R1 + R2 + R3 I0 = Replace the part of the circuit that contains I1 and R1 with its Thevenin equivalent; then do the same with the part that contains I2 and R2 : R1 Io R3 R1 I1 R2 I2 Now the resistors are in series, as are the voltage sources: Io = R2 R1 I1 − R2 I2 R1 + R2 + R3 11 8 Poles (10 points) Each signal below has the form s[n] = (a + bj)n + (a − bj)n where a and b can have values 0, 0.3, 0.5, 0.9, 1.1, −0.3, −0.5, −0.9, −1.1. The periodic signals have a period of either 2, 4, or 8. For each one, specify a and b. 2.0 2 2.0 1.5 1.5 1 1.0 1.0 0.5 2 2 4 6 8 10 4 6 8 10 0.5 -1 -0.5 2 a : −0.3 b : 0 a : −0.9 b : 0 4 6 8 10 8 10 8 10 a : 0.3 b : 0 2.0 1.5 10 1.0 5 4 2 2 0.5 2 2 4 6 8 10 4 2.0 4 1.5 3 1.0 2 0.5 1 8 10 4 2 4 6 6 -0.5 4 2 8 a : −1.1 b : 0 2.0 1.5 1.0 0.5 a : 0.5 b : ±0.5 a : 1.1 b : 0 6 -4 2 6 10 a : 0.9 b : ±0.9; 5 4 8 -5 a : 0.9 b : 0 2 6 4 -2 10 -2 -4 a : 0.0 b : ±1.1 12 8 10 -0.5 -1.0 -1.5 2 4 6 a : 0.0 b : ±0.9